Ball A with a mass of 0.500kg is moving east at a velocity of0.800m/s. It strike

Question

Ball A with a mass of 0.500kg is moving east at a velocity of0.800m/s. It strikes Ball B, also of a mass of 0.500kg, whichis stationary. Ball A glances off B at an angle of 40 degreesNorth of its original path. Ball B is pushed along a pathperpendicular to the final path of Ball A.
a) What is the momentum of ball A after the collision?
b) What is the momentum of ball B after the collision?
c) What is the velocity of ball A after the collision?
d) What is the velocity of ball B after the collision?

Explanation / Answer

Applying the law of conservation of momentum along x-axis wehave as, m1u1x+m2u2x=m1v1x+m2v2x 0.5*0.8+0.5*0=0.5*vcos60+0.5*ucos30 vcos60+ucos30=0.8 v+3u=0.8 Along y-axis we have as, 0.5*0+0.5*0=0.5*vsin60+0.5*-usin30 vsin60=usin30 3v=u Solving we get 4v=0.8 v=0.2m/s velocity of ball A u=0.3464m/s velocity of B Its momentum along along x-axis of ball A is px=0.5*vcos60 along y-axis of ball A is py=0.5*vsin60 resultant p=px2+py2 at an angle tan=py/px with respect to x-axis. Similarly for the ball B. Hence we get by it. at an angle tan=py/px with respect to x-axis. Similarly for the ball B. Hence we get by it.

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